Step 1

Given that W is the subspace of all diagonal matrices in \(M_{2,2}\)

The objective is to a basis for W.

Step 2

Consider the given vector space W of all diagonal matrices in \(M_{2,2}\)

Therefore,

\(W=\left\{\begin{bmatrix}a & 0 \\0 & b \end{bmatrix}, \text{a and b can be any real number}\right\}\)

Let E be basis for W.

Therefore,

\(\begin{bmatrix}a & 0 \\0 & b \end{bmatrix}=a\begin{bmatrix}1 & 0 \\0 & 0 \end{bmatrix}+b\begin{bmatrix}0 & 0 \\0 & 1 \end{bmatrix}\)

Now, let \(\lambda_1\) and \(\lambda_2\) be any two scalars such that:

\(\lambda_1\begin{bmatrix}1 & 0 \\0 & 0 \end{bmatrix}+\lambda_2\begin{bmatrix}0 & 0 \\0 & 1 \end{bmatrix}=[]\)

\(\begin{bmatrix}\lambda_1 & 0 \\0 & 0 \end{bmatrix}+\begin{bmatrix}0 & 0 \\0 & \lambda_2 \end{bmatrix}=\begin{bmatrix}0 & 0 \\0 & 0 \end{bmatrix}\)

\(\begin{bmatrix}\lambda_1 & 0 \\0 & \lambda_2 \end{bmatrix}=\begin{bmatrix}0 & 0 \\0 & 0 \end{bmatrix}\)

Equating the elements:

\(\lambda_1=0 ,\lambda_2=0\)

Therefore, \(\begin{bmatrix}1 & 0 \\0 & 0 \end{bmatrix}\) and \(\begin{bmatrix}0 & 0 \\0 & 1 \end{bmatrix}\) are linear independent.

Hence, the basis of W is \(E=\left\{\begin{bmatrix}1 & 0 \\0 & 0 \end{bmatrix},\begin{bmatrix}0 & 0 \\0 & 1 \end{bmatrix}\right\}\) and dimension is 2.

Given that W is the subspace of all diagonal matrices in \(M_{2,2}\)

The objective is to a basis for W.

Step 2

Consider the given vector space W of all diagonal matrices in \(M_{2,2}\)

Therefore,

\(W=\left\{\begin{bmatrix}a & 0 \\0 & b \end{bmatrix}, \text{a and b can be any real number}\right\}\)

Let E be basis for W.

Therefore,

\(\begin{bmatrix}a & 0 \\0 & b \end{bmatrix}=a\begin{bmatrix}1 & 0 \\0 & 0 \end{bmatrix}+b\begin{bmatrix}0 & 0 \\0 & 1 \end{bmatrix}\)

Now, let \(\lambda_1\) and \(\lambda_2\) be any two scalars such that:

\(\lambda_1\begin{bmatrix}1 & 0 \\0 & 0 \end{bmatrix}+\lambda_2\begin{bmatrix}0 & 0 \\0 & 1 \end{bmatrix}=[]\)

\(\begin{bmatrix}\lambda_1 & 0 \\0 & 0 \end{bmatrix}+\begin{bmatrix}0 & 0 \\0 & \lambda_2 \end{bmatrix}=\begin{bmatrix}0 & 0 \\0 & 0 \end{bmatrix}\)

\(\begin{bmatrix}\lambda_1 & 0 \\0 & \lambda_2 \end{bmatrix}=\begin{bmatrix}0 & 0 \\0 & 0 \end{bmatrix}\)

Equating the elements:

\(\lambda_1=0 ,\lambda_2=0\)

Therefore, \(\begin{bmatrix}1 & 0 \\0 & 0 \end{bmatrix}\) and \(\begin{bmatrix}0 & 0 \\0 & 1 \end{bmatrix}\) are linear independent.

Hence, the basis of W is \(E=\left\{\begin{bmatrix}1 & 0 \\0 & 0 \end{bmatrix},\begin{bmatrix}0 & 0 \\0 & 1 \end{bmatrix}\right\}\) and dimension is 2.